Internat. J. Math. & Math. Si.
Vol. 5 No. 4 (]982) 691706
691
A POLYA "SHIRE" THEOREM FOR FUNCTIONS
WITH ALGEBRAIC SINGULARITIES
J.K. SHAW and C.L. PRATHER
Department of Mathematics
Virginia Polytechnic Institute and State University
Blacksburg, Virginia 24061-4097
(Received February 5, 1982)
ABSTRACT.
The classical "shire" theorem of
braic poles, in the sense of
bralc pole at z
0
P61ya
L. V. Ahlfors.
is proved for functiuns with alge-
A function f(z) is said to have an alge-
provided there is a representation f(z)
lk=_Nak(z
where p and N are positive integers and A(z) is analytic at z
O.
given reduces to an entirely new proof of the shire theorem.
For p
z
+ A(z),
0)
i, the proof
New quantitative results
are given on how zeros of the successive derivatives migrate to the final set.
KEYWORDS AND PHRASES.
Final set, algebraic singuly, branch line, shoe theorem.
1980 MATHEMATICS SUBJECT CLASSIFICATION CODE.
i.
Primay 50A08; Secondary 30A68.
INTRODUCTION.
Plya [1,2]
follows.
A point
defined the "final
z0eS(f)
many derivatives of f(z).
set" S(f) of
a meromorphlc function f(z) as
if every neighborhood of z contains zeros of infinitely
0
The final set determines, roughly speaking, the final
position of the zeros of the derivatives of f(z).
For functions with two or more poles,
showing that
zoeS(f)
if and only if z
0
Whlttaker’s "shire" [3] descripition of
Plya
characterized the final set by
is equidistant from two or more poles.
P61ya’s
Theorem is quite illuminating.
Let us
J.K. SHAW AND C.L. PRATHER
692
say that the shire of a pole I of f(z) will consist of all points z in the plane which
are nearer to I than to any other pole of f(z).
Then the final set of a meromorphic
function consists of the boundaries of the shires of its poles.
If f(z) has but one pole, the final set is empty, and the single pole generates
zero-free regions for the successive derivatives of f(z).
The asymptotic size of such
Thus the poles of f(z) may be thought of
zero-free regions has been obtained in [4].
as repellers of equal strength of the zeros of the derivatives of f(z).
of the "equal strength" aspect of the
Plya
A refinement
shire theorem is one of the features of
this paper.
The general questions posed in P61ya’s classic paper [i] concern the effect of
the singularity structure of an analytic function on its final set.
The present paper
takes up the problem of determining the final sets of functions with algebraic singu-
laritie8.
We take a countable set A of points in the plane
(%) denote the "principal branch line" (%)
{z
We assume in addition that each bounded region D
tlt
+
%
,
<
and for each lea we let
0}, corresponding
to I.
intersects the branch lines of
The functions f(z) we study are those which have "algebraic
only finitely many leA.
poles", in the sense of Ahlfors [5, p. 299], at the points of A.
To say that f(z) has
an algebraic pole at z means that there are integers p > i and N > i such that f(z)
0
has the representation
[ Ck(Z-Z O) k/p
k=-N
f(z)
in a neighborhood of z
coefficients
A(D)
Ck(1),
{leAl(1)
0.
-N(1)
For each lea we associate integers p(1)
<
k <
,
.
with
D}; A(D)
intersects
tion
f(z)
[
if
Ira(z)
By (z
l)
Im(1) and Re(z)
over into the analytic part
>
C_N(l
(1) # 0.
is a finite set.
lea (D)k=-N(1)
is analytic in D.
(1.1)
C_N#0
Ck(1)(z -l)
>
i, N(1) > i and
Given a bounded region
We shall suppose that the func-
k/p(1)
AD(Z),
The factors (z-l)
(1.2)
(z)
we mean the branch which is analytic for
Re(1).
D, let
k/p(1)
k > 0
z(%)
and real
cannot be thrown
as would happen if f(z) were meromorphic.
Thus the
FUNCTIONS WITH ALGEBRAIC SINGULARITIES
693
singularities of f(z) in D arise from the algebraic poles and branch lines there, and
may be subtracted off in the manner of poles for meromorphic functions.
Since the
unique representations (1.2) for f(z) are to hold for arbitrary D, then the coeffi-
k()}k= 0 must be entire coefficients; i.e., for each
k0Ck(%)(z %)k/p(%) must converge for all complex numbers z.
cients {c
%EA, the series
Disregarding the branch lines, form the shires of the points %cA by the classical
definition and let P(A) be the union of all the boundaries of the shires.
Our princi-
pal result asserts that in spite of algebraic singularities, P(A) is still the final
set of
f(z),
with one exceptional case.
Not surprisingly, difficulties arise when a
horizontal line segment in P(A) determined by two points
branch line of a third point
are of
unequal strength,
(N2/P2),
where N
THEOREM i.
1
%3eA.
Pl
coincides with the
Should this occur, we will suppose that
in that either
N(% I),
%l,%2eA
P(%I )’
(Nl/Pl)
#
(N2/P2)
or
feN1 ICN21
#
II
if
and
2
(Nl/Pl)
etc.
If f(z) is defined and restricted as above, the final set of f(z) is
Note that the theorem makes no mention of the branch lines apart from the exceptional case; the branch lines come into play only in the proof, and only in the excep-
tional case.
We prove Theorem i in the next section, after which we make a series of Remarks
on how the proof can be modified to cover various situations.
For example, it is
reasonable to ask how the theorem would be altered if we allow zeros of derivatives of
all branches to contribute, instead of a single, fixed branch.
There is no hope of treating the case where (1.2) is replaced by a doubly infinite series.
Even if p
i, essentially nothing is known about final sets for
functions with isolated essential singularities.
Our proof of Theorem i covers the case where f(z) is meromorphic (p--i) and without restriction on the location or strength of poles. As such, it is a new proof of the
Plya
shire Theorem, independent of the two existing proofs in
[6], [I], [3].
More-
over, the classical proofs do not extend to the case of algebraic singularities, and
this is why it was necessary to devise a new proof.
J.K. SHAW AND C.L. PRATHER
694
The classical theory asserts that the location of the final set does not depend
on the order of the poles or the size of their coefficients.
Even so, our proof
reveals the previously unobserved phenomenon that if adjacent poles have unequal
strength, then all the zeros of sufficiently high order derivatives are pushed to one
side of the common boundary of the two shires.
Various other features of the proof bearing on the migration of zeros to the flna.
set are mentioned in the Remarks.
2.
PROOFS AND REMARKS.
It is convenient to express the derivatives of the factors (z
the Gamma function, and in terms of certain factors
paper [4].
where p > i, so that the
n
dz
n
(z
in terms of
which were studied in the
For an arbitrary integer k, let us define
k(k + p)(k + 2p)
d
Ck,n
%)k/p
%) k/p
tn-h derivative
%)k/p
of (z
l)p)
may be written as
.((k/p)-(n-l)) (z-%) (k/p)-n
(k/p) ((k/p)-l)
(-l)
(k + (n
np-n (-k) (-k + p) (-k + 2p)... (-k+(n-l)p) (z-%) (k/p)-n
(2.1)
-n
(-i) n p
C_k,n(Z-
%) (k/p)-n
(-l)n{F(-kp -I +n)/F(-kp -I) }(z-%) (k/p)-n
(see [7, p. 25.-,).
We require certain information about the asymptotic behavior of
the terms in (2.1).
To begin with, if a and b are arbitrary real numbers, then [7, p.
257]
li
n-=
{nb-aF(a + n)/F(b + n)l
Therefore, if N is a positive integer
CI ,n
CN,n
F
(p-l+n)
r(p -I
F (Np-.
we have
I)
r (Np-l+n)
using order-of-magnitude notation.
1
0
(n(1-N)/P),
n
=,
FUNCTIONS WITH ALGEBRAIC SINGITIES
695
Let [x] denote the greatest integer not exceeding the real number x, and define
max
l_<k<
(n-l) p]
IC-k,n/Clnl
n-- 1,2,3,...
We shall need the result of Lemma 3.1 of [4] which asserts that there is a constant A,
which depends only on p, such that
6
n
n
I/p -< A
(2.2)
i, 2, 3
n
Finally, we note that if k > [(n-l)p] +i, then p > i implies k > (n-l)p > n-l,
Also for k > [(n-l)p] +i,
and therefore k > n.
k(k-p)...(k-(n-l)p)
k!
n! (k-n)
Cln
<
k!
n! (k-n)
(k- (n-l))
k(k-l)
(1)(2)---(n)
(i) (l+p).-. (l+(n-l)p)
(2.3)
--( kn
the binomial coefficient.
Our first lemma is a result which shows that compact subsets of shires of points
A similar result
are zero-free for sufficiently high order derivatives of f(z).
for meromorphic functions appears in Polya’s original proof [1,3].
LEMMA I.
Let aeA and let f(z) be defined as in Theorem i.
lid
z
If (n)<-)/n!l lln
al
Then
1
uniformly on compact subsets of a-shire.
Let K be a compact subset of a-shire and let R > 0 be such that K is con-
PROOF.
tained in
Iz
intersects
as
f(z)
a
D,
.
k---N
where
(z)
<
(R/2).
# a}.
Then for zD-
ak(z-a)
Iz
%AU(D)(%)
Let D be the disc
k/p +
A(D)
.
a
and
<
R and put A(D)
z(a),
cj(%)(z-%)
j =-N (k)
j/p(E)
is analytic in D and where we have written a
i
{%AI(%)
f(z) can be represented
+
(2.4)
(z)
ci(a)
and N
N(a).
J.K. SHAW AND C.L. PRATHER
696
.
Computing derivatives in (2,4) and using
(_l)nf (n) (z)
k=-N
obtain
{akF(_kp-l+n)/F(_kp-l)l(z_a) (k/p)-n
leA(D)
+
(2.1), we
j=-N(1)
(2.5)
(-l)n4n) (z)
a_NF (Np-l+n)
{i
(Nlp)+n
-I)
(z-a)
F (Np
+ Fn (z) + Gn(z) +
Hn(Z)
where
[,
F (z)
n
k=-N+l
G (z)
n
leA(D)
akr (Np -I) r (-kp-l+n) (z a) (k+N) /p
a_NP (-kp -I) r (Np-l+n)
c, (1) F (Np-I) F (-j p (1)-l+n)
j=-N(1) a_NF (-jP (1)-l) F (Np-i +n)
F(Np-I) (z-a) (Nlp)+n
a_Ni, (Np-l+n)
Hn (z)
unifory on D.
Starting with F
n
(z),
greatest integer function) and [(n-1)p]
+ 1
Gn(Z)
+ 1
k <
first range has order of magnitude, for large n,
=,
since k >- -N
+
i.
0 vanish since i/r(0)
and H (z) all converge to 0
n
.
-1,
k
1
k
[(n-1)p]([x]
The kt--)-h term of F (z) in the
n
r(-kp-l+n)/r(Np-l+n)
0(n-k+N)/p)
Since the sum over the first range has only finitely
many terms, it follows that this part of Fn (z) is uniformly small for large n, zeD.
The sum over the second range is
in view of (2 i) and the definition of 6 n
by
la_N i-i
_<
J
(n-l)p]
k=l
a-Nl- I
lakl IC_k,n/Cl n[
[(nl)p]..lakl
k=l
0.
we first break its defining sum into
three parts, with k ranging successively over -N
o(i), n
(z_a)- (N/p)-n
0 and j
We are going to show that the sequences Fn (z)
,
(j Ip (1))-n
(-l)nn) (z)
For n > 0 the terms in (2.5) corresponding to k
as n
(z-h)
6n n
[C I ,n /CN ,n
(1-N)/p
]z
[z
al (k+N)/p
aI(k+N) Ip
bounded
FUNCTIONS WITH ALGEBRAIC SUNGULARITIES
By (2.2), the above does not exceed
where J is constant.
JAIa-N I-In-(N/p) k=l lakl Iz
Since
akl
k=l
nent of F
n
697
z
al(k+N)/p"
k/p
a
D, it follows that the compo-
is uniformly bounded over
(z) coming from the sum over the range i
_<
k -< [(n
l)p] + i, we use (2.1) and
Looking at the third range, k -> [(n
uniformly to 0 on D.
l)p] also converges
(2.3) and bound the sum by
J
J
lak ll/k
Since
la_N I-I n(I-N)/P
k=n
.01a_NI-I n(i-N)/Plz-a IN/p k=n
O, then
I/p
lak ll/k Iz- al
<-y
(1/2) for all zeD, if k is
<
not exceed
k
yn
k=n
for n sufficiently large.
the range k > [(n-l)p]
.
(nk)( lakil/klz_all/p)k
-
This means that the infinite sum in the above expression does
sufficiently large.
n
Iz al (k+N)/p
nk
lakl
+
Since
k
Y
(/(l-y))
-i
i
n+l
< i, then the component of F
n
i is asymptotically smaller than
This completes the proof that F (z)
n
0, n
(z) taken from
n(l-N)/P(/(l-l)) n+l
O,
uniformly in D
As for G (z), we need the fact that K is compact and in a-shire so that there
n
assuredly is a number
leA(D).
0 <
T
<
i, such that
z
a
z
<
l
for all zeK and
We pick one leA(D) and break the corresponding sum over j into three parts
exactly as above.
of magnitude
"r
n
<_ j <
A typical term in G (z) lying in the range -N(1)
n
n-(J+N)/P(1)
formly to 0 over K.
+ I
T,
,.
j _<-i has order
Hence this component of G (z) tends unin
The arguments for the ranges i <_ j _<
[(n-l)p(1)] and [(n-l)p(X)]
run exactly as the corresponding ones for F (z) except that we now have
n
zeK and the additional factor
that G (z)
n
O, n
_<
0, n
n before each sum.
, uniformly
on K.
Working with Hn (z), we first have
We omit the details but conclude
J K. SHAW AND C.L. PRATHER
69 8
a) n
(z
n!
gn)(z)
n’
a) n
(z
.laj+nl0
j+n
a) n
0(i), n + j
n)(z)
If zeK then
z
0
(Iz -allo)
<-7
+n
m, if
j +n
0 <
+
n
(i12), for some
n
0
n
[ (k)
ek(z-a)
n
k= n
uA(z)
z
a
is analytic in D.
j
Therefore
0
0 < R and a constant 7, and so the
y))n.
0-term in (2.6) is dominated by O(y/(l
H (z)
a)k-n
a) j+n
(z
R, since
j
a
<
k!k
(k-n)’ (a
k=n
n
j--0
Now
k=0
[
n’
j
a)k
[ ek (z
n
dz
Then since n!
1-7
F(n + i), we have
zeK
o(i), n
Going back to (2.5) we now have
(z-a) (Nlp)+nf (n) (z)
n!
and since
[r(Np-l+n)Ir(l+n)]
lln
a-Nr (Np_l+n)
r(Np -I) r (n+l)
n
(N/p)-I
(1 + o(1)
lln
the desired result follows
Lemma i shows that no point of any shire can be in the final set.
The next lemma
establishes the existence of zeros of derivatives near points on the boundaries of
shires, and thus Theorem i will follow.
LEMA 2.
borhood of
PROOF.
Let $eP(A) and let f(z) be defined as in Theorem i.
$ contains zeros of
f(n)(z)
for n sufficiently large.
First we note some reductions.
If $ lies on the boundaries of a-shire,
b-shire and other shires, then we can consider a point
boundaries of a-shire and b-shire only.
’
near
which lies on the
If we produce zeros in arbitrary neighbor-
hoods of $’, then zeros in neighborhoods of
We note that in this reduction the point
Then every neigh-
’
$ are implied by elementary arguments.
can also be taken not to lie on any branch
699
FUNCTIONS WITH ALGEBRAIC SINGULARITIES
line with the single exception that a and b have the same real part.
In this case, re-
Our proof, then, splits in-
call our hypothesis that a and b are of unequal strength.
either does or does not lie on a branch line, but in any case
to the two cases where
is on the common boundary of exactly two shires.
CASE I:
E does not lie
on a branch line.
Suppose
I al
< R and
Iz b
ze- hA)(%),
of a-shire and b-shire, let R > 0 so that
Iz
be the union of the discs
sects
,
%
# b}.
# a,
a
Then if
E
lies on the common boundary
<
(R/2) and
<
R, and put
b
<
(R/2), let
A()--{eAl(%)
inter-
f(z) may be expressed as
Z ak(z-a) k/P + k=-M bk(Z-b)k/q
f(z)
k---N
[
+
[
%eA()
where
A(z)
and p,
is analytic in
j=-N(%)
ak, N,
q,
bk,
cj()
(z-) jlp() +
A(z)
M have obvious meaning.
If we
differentiate this representation, we get
f(n) (z)
[ (-l)n [akF (-kp-l+n) / F(-kp -I)
(z-a)
n[bkF (-kq-l+n) / F (-kq -I)
(z-b)
(k/p)-n
k=-N
[
+
(-i)
k=-M
+
+
(-1)n [cj () F (-jp ()-l+n)/F(-jp()-i) (z_%)(j/p (%))-n
Z
Z
%eA()j=-N(%)
(n)
A
(z)
(-1)
na_NF (Np-l+n)
+
F(Np-z-a)(N/p)+n {i Fn(Z)
where
F (z)
n
G (z)
n
[.
k---N+l
Z
(k/q)-n
k---M
akF (-kp-l+n) F (Np -I) (z-a)
+ Gn(z) +
Hn(Z)
+ I
n(z)}
(k+N)/p
-1
a_NF (-kp F (Np-l+n)
(N/p)+n
DkF (-kq-l+n) F (Np -I) (z-a)
-/q)+n
-i
a_NF(-kq )F(Np-l+n) (z-b)
(2.7)
700
J.K. SHAW AND C.L. PRATHER
c. (%) F (_j p (%)
n
a_NF(-jp(%)-l)F(Np-l+n) (z-%) -(j/p())+n
%eA() j=-N(%)
(-I)nF(Np-I) (z-a) (N/p)+n A(n) (z)
-I
a_NF (Np + n)
I (z)
Gn(Z)
If we group
-l+n) F (Np-l) (z_a) (N/p)+n
with I (z) and factor out the term
n
[(z-a)/(z-b)]n
we get the ex-
pression
+
G (z)
n
bkF (-kq-l+n) F (Np -I)
[
In(Z)
a_Nr (-k q -i) r (Np-l+n)
k=-M
(z_a) N/P (z_b)
(z-a) N/p (z-b) n A (n) (z)
+ (-I) nF (NP-I) -I
a_Nr(Np + n)
[Un(Z)+
V (z)]
n
(’z a)
z
z
,
T
n
a
<
(R/2)} u
{[z
b[
n
n
=[[a-b-MF(Mq-I
.(Np
-I
+ n)
F(Np -I) (z
+ n)
F(Mq-l)(z
-i
+
k-M+l
(R/2)}
<
We calculate the asymptotic form of U (z) as n
U (z)
Let T be a compact
in its interior and
with
{Iz
b
b
where U (z) and V (z) denote the respective terms in the brackets.
n
n
neighborhood of
k/q
,
%eAu()(%)
Write U (z) in the form
n
zeT.
N/p
a)
M/q
b)
+
-I
bkF(Np )F(-kq + n)
-1
b_r (-kq -1) r (Mq + n)
N/p(z b)k/q
].
We proceed to treat the above infinite sum inside the brackets just as in Lemma i
i.e., we break it up into summands with ranges
[(n-l)q] + i
-(k+M)/q
n
k <
0, n
.
,
The
kth term
since k e -M
+
M
k
-i, i
k
[(n-l)q] and
in the first range has dominant asymptotic form
i.
In the middle range, take note of (2.1) and
(2.2) to see that the significant terms asymptotically are
701
FUNCTIONS WITH ALGEBRAIC SINGULARITIES
(n-1)q]
[bk[ [C-k,n/Cl,n[ [Cl,n/CM,n[ [z b[k/q
k--i
(n-l)q]
<
[bk[ 6n n (I-M)/q
J
k=l
<
JA(q) n -M/q
k=l
b[k/q
z
[bk[ [z b[k/q
where J is a constant and the terms of (2.1) now depend on q instead of p.
tends to 0 with n
on T.
Ek= I [bk[ [z-b[
in view of the fact that
The sum in U (z) over the range [(n-1)q]
n
k=
.
+ i
k <
(l-M)/q
k
Z
k--n
bk[
n
and the rest of the proof runs as in Lemma i.
n(z)
U
is uniformly bounded
z
b[ k/q
a[k/q
z
Consequently, we may say that
r(Np-1) (z.-.a)N/P
F(Mq -I) (z-b) M/q
a_Nr(Np-l+n
This term
behaves llke
[bkl IC_k, n /Cl,nl ICI, n/CM,n[
(n-l) q]+l
<Jn
k/q
+ o(i)
n
where o(i) denotes a term which,converges uniformly to 0 in T.
Since T lies in
tion
AD(Z)
in
[z
(2.6).
b[
we may treat the term V (z) as we did the funcn
That is,
(n)
]a_Nr
r(Np-1(Np-l+n)
.).r(l. +n)
IVn(Z)[
0
which means that V (z)
n
(R/2),
<
n
1-(N/p
A
(z)(z
y
i’
"Y
n
n
In(Z)
zeT
Moreover, if we let Wn (z)
it then follows that
+
N/p
n
0 uniformly in T.
Gn(Z)
b) n
z-an
Wn(Z) (z
b
U (z) + V (z),
n
n
J.K. SHAW AND C.L. PRATHER
702
where
wn (z)
lira
n-= n
(M/q)-(N/p)
--W
(z-a) N/p (.W-- constant)
M/q"
(z-b)
and the convergence is uniform in T.
The functions F (z) and H (z) in (2.7) have the same form as F (z) and G (z) in
n
n
n
n
It is easy to see that the arguments of Lemma i carry over to show that F (z)
n
(2.5).
0 in (2.7), uniformly in T.
and H (z)
If we let
f (z)
n
then f (z)
0 if and only if
n
(-l)nr (Np -I) (z-a) (Nlp)+n
a_Nr (Np
f(n)(z)
-i
f
(n)
(z)
zT
+ n)
0, and the foreBoing discussion yields the
representation
f (z)
n
I + Fn (z) +
Hn(Z)
+
z-a)n
Z b
Wn(Z)
(2.9)
where (2.8) holds, and moreover,
F (z) + H (z)
o(i),
n
n
(2.10)
zeT
n
All terms in (2.9) are analytic in T.
.
(z-a)/(z-b) for z
Let us now consider the covering properties of the mapping w
near
e
lies on the perpendicular bisector of the line segment from a
Moreover, we may assume without loss of generality that
to b.
m
Recall that
i0
# (a+b)/2.
under the map lies on the unit circle in the w plane.
of
of the perpendicular bisector is the circle
lwl
The image
In fact, the image
Note that m # I and m # -I
i.
(the image of (a + b)/2) so that we can surround m with a sectorial neighborhcod
S
r <
]w]
< r
the points w
is the region
-1,
0
i and w
S
g < arg w <
-i.
+ g,
where 0 < r < 1 and g > O, whose image omits
The inverse image of S
in the z-plane which contains
which are the inverse images of the circle
Iwl
under the mapping w= (z-a)/(z-b)
and is bounded by the four circles
r,
Iwi
r
-1
and lines arg w
+
,
FUNCTIONS WITH ALGEBRAIC SINGULARITIES
e
arg w--
6; see Figure i.
Let F I,
F2’ 3’ 4
respectively by appropriate portions of
+ 6,
and let
S.
aries of
YI’ 2’ Y3’ 4
lwl
703
denote the boundaries of S
r, arg w
e
lwl
6,
r
formed
-I and
arg w
be the corresponding inverse images forming the bound-
r) are sufficiently small that
and (i
Finally, suppose that
S
T.
Figure i
Let e >0.
By (2.8) we can make the argument of W (z),
n
vary by less than
n
by taking n sufficiently large; i.e., A arg
T
Now as z ranges over
r.
by taking n large.
W
(z)[(z-a)/(z-b)]n
Ay2
for
< e for all
Y2
and
Y4’
w
71’
can be made large.
ze71
>
+ 4e, for
4n
Specifically
we require
all large n.
(z-a)/(z-b) has .constant argument, so we may impose
n(z)[(z-a)/(z-b)]n
arg W
<
and
n large enough.
On
Wn(Z)
Thus for suitably large n, the variation of the argument of
Ayl Wn(Z)[(z-a)/(z-b)]n
On
z ranges over a small T,
(z-a)/(z-b) traces out the circular arc F on
1
n
Thus we can make [(z-a)/(z-b)] wrap around lwl
r as many times as desired
large n.
lwl
as
71’ l(z-a)/(z-b)l
A4
arg
Wn(Z)[(z-a)/(z-b)]n
<
simply by taking
whereas W (z) has order of magnitude n
n
r < i
by (2.8).
Thus if n is large enough,
Similarly,
IWn(Z)[(z-a)/(z-b)In
>
IWn(Z)[(z-a)/(z-b)]nl
(5/4) for all
zo
<
(M/q)- (N/p)
(3/4) for all
when n is large.
ZeTl.
J.K. SHAW AND C.L. PRATHER
704
Taking into account the arguments and modull restricted as above, we see that the
S under
image of
(3/4)
lwl
<
<
the map
Wn(Z)[ (z-a)/(z-b)]n
contains the doubly-covered annulus
Hence the image of T under W (z)[(z-a)/(z-b)]n contains a disc
n
(5/4).
centered at -i of radius (1/4) (in fact, the image contains a disc of radius (1/4)
centered at each point of
I).
w
Fn(Z)
All that remains is to make n large enough that, say
INn (z)
Rouch’s
Then by (2.9) and
(i/16)for zeT.
<
(1/16) and
r
f (z) maps
Theorem
region which contains a disc centered at 0 of radius (1/8).
<
n
onto a
Thus f (z) has zeros in
n
T for all n sufficiently large, and this finishes Case I.
lies on a branch line.
CASE II.
lies on the common boundary
We suppose that
We represent f(z) exactly as in
Re(b).
of a-shire and b-shire only, where Re(a)
Case I, where this time at least one %eA() has imaginary part (a
+ b)/2.
By
hypothesis, a and b have unequal strength, and we assume without loss of generality
that b is stronger than a; i.e
-I
either Mq
>
Np
-I
b_M
or
>
if Mq
a_N
.
-I.
-I
NP
In (2.9) it is no longer true that each term is analytic in a neighborhood of
for at least one term in the sum H (z) has a branch line through
n
terms in
S
(2.9) are analytic in the lower half
of
S,
However, all
and it is easy to see that
S.
(2.8) and (2.10) hold uniformly over
,
.
We now proceed with a covering type argument nearly identical to Case I, but for
S.
the half neighborhood
Y2
and
Thus,
Y4’
,
and let
7, 7,
7
Let
we make
form the boundary of
enough.
Wn(z)
AS argWn(Z)
and
7
be the lower halves of the arcs
be the segment of the perpendicular bisector joining
parts of the boundary of the image
The function
7
let
i’
S*
of
is analytic on
< e
and A
and
S. Let F, F, F, F be the corresponding
S under w (z-a)/(z-b).
S and continuousn on
arg Wn (z)[(z-a)/(z-b)]
>
its closure.
,
As before,
4 + 4e by choosing n large
Moreover, (z-a)/(z-b) still has constant argument on
Ay,argWn(Z) [(z-a)/(z-b) in
y
7
and
and therefore
A i.argW (z) [(z-a)/(z-b) ]n < e if n is taken large
n
< e
y
enough.
As for what happens on
from U (z).
n
If Mq
-I
>
Np
-I
7,
we consult
then W (z)
n
(2.8) and note the constant W which comes
,
n
,
everywhere on the closure of
S*.
705
FDCTIONS WITH ALGEBRAIC SINGULARITIES
-I
If Mq
Np
-I then
Ib_M
>
la_N
and it follows that
and so all we can say is that for some constant n > 0,
for all
z
zes", when
all
n
lWn(Z)[(z-a)/(z-b)]n
Consequently, the image of
n is large.
[(z-a)/(z-b)]n
W (.)
As before,
and n sufficiently large.
Wn(Z)[(z_a)/(z_b in
>
< i
i
+
n for
S under the mapping
contains a doubly covered annulus 1
n
<
lwl
1
+ n. In par-
ticular, the image contains a disc centered at -1 of radius
Now we make
large.
Then
IFn(Z)
fn(Z)
i
<
+
IHn(Z)
(n/4) and
Fn(Z)
+
Hn(Z)
+
<
(n/4) for
zS and n sufficiently
Wn(Z)[(z-a)/(z-b)]n maps S over a disc
centered at 0 of radius (n/2) which completes the proof in Case II.
This completes the proof of theorem i.
I.
If f(z) is meromorphic then a trivial modification of Case I gives a proof of the
Polya shire theorem, without restriction on the location or strength of poles.
argument shows, moreover, that we can make the image
wrap around the annulus (3/4) <
large enough.
lwl
<
(5/4), any given number of
(z-a) / (z-b) in
times by taking n
Having specified a positive integer k, then, all the derivatives beyond
a certain point will have at least k zeros in
from
S under Wn(Z)
The
(a-b)/2, the further
is away from-i.
S.
Note that the further away
Loosely speaking,
have to be taken relatively larger in order that the image of S
W (z)[(z-a)/(z-b)]
n
n
contain the point -i.
his
is
means n will
under the map
In otherwords, the further out a point is
from the poles, the longer one has to wait for the appearance of zeros of derivatives.
Consider a meromorphic function with poles a and b, where b is stronger than a.
S* of S is pushed out
< i by Wn(Z)[(z-a)/(z-b)]n. Therefore
S* will be free of zeros
Arguing as in Case II, we see that the image of the upper half
away from the disc
of
f(n)(z)
lwl
for n large enough.
than pole a, pushes the zeros of
Put another way, we may say that pole b, being stronger
f(n)(z)
past the perpendicular bisector, even though
the cluster points of zeros remain on the bisector.
Note that meromorphic functions
real on the real axis, whose derivatives have zeros occurring in conjugate pairs, have
poles which occur in conjugate pairs, with the same orders, and with conjugate Laurent
coefficients.
J.K. SAW AND C.L. PRATHER
706
2.
The choice of the principal branch in Theorem I was only for convenience.
Any
fixed branch of f(z) would serve as well, with minor modifications in the proof.
We
would have to retain the hypothesis that any bounded region intersects only flnitely
many branch lines.
3.
If f(z) has a single algebraic pole, the final set is empty.
This is a special
Moreover, the results of [4] regarding the asymptotic size of
instance of Case I.
zero-free regions created by a single singularity hold with minor modifications.
4.
If we alter the definition of the final set so as to allow all zeros of all
derivatives of all branches, then the conclusion of Theorem I holds without restriction
on the location or strength of singularities.
Even if infinitely many branch lines
coincide with a perpendicular bisector, we may say that there exists a branch of f(z)
whose branch lines are elsewhere.
Hence Case II is entirely avoided.
REFERENCES
I.
Uber die Nullstellen Sukzesslve Derivierten, Ma. Zi. 12 (1922),
POLYA, G.
36-60.
2.
POLYA, G.
3.
WHITTAKER, J.M.
4.
SHAW, J. K. and PRATHER, C.L.
5.
AHLFORS, L.
Complex Analysis (third edition), McGraw-Hill, 1979.
6.
HAYMAN, W.K.
Meromorphic Functions, Oxford University Press, 1964.
7.
ABRAMOWITZ, M. and STEGUN, I. (Eds.)
On the zeros of the Derivatives of a Function and its Analytic Char(1943), 178-191.
acter, Bu. A...
Interpolatory Function Theory, Cambridge University Press, 1935.
.
Zeros of Successive Derivatives of Functions
2 (1982),
Analytic in a Neighborhood of a Single Pole, ha
111-119.
a.
Handbook of Mathematical Functions with
Graphs andMathematical Tables, National Bureau of Standards
Appl. Math. Ser., No. 55, uerlntenden-of Documents, U.S. Government
Printing Office, Washington, D.C., 1964. MR29#4914.
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